Lösung 2.2:2c
Aus Online Mathematik Brückenkurs 1
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- | We can simplify the left-hand side in the equation by expanding the squares using the squaring rule | + | We can simplify the left-hand side in the equation by expanding the squares using the squaring rule |
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+ | {{Displayed math||<math>\begin{align} | ||
+ | (x+3)^{2}-(x-5)^{2} | ||
+ | &= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt] | ||
+ | &= x^{2}+6x+9-x^{2}+10x-25\\[5pt] | ||
+ | &=16x-16\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
Thus, the equation is | Thus, the equation is | ||
+ | {{Displayed math||<math>16x-16=6x+4\,\textrm{.}</math>}} | ||
- | + | Now, move all ''x'''s to the left-hand side (subtract 6''x'' from both sides) and the constants to the right-hand side (add 16 to both sides) | |
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- | Now, move all | + | |
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- | from both sides) and the constants to the right-hand side (add | + | |
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- | to both sides) | + | |
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+ | {{Displayed math||<math>\begin{align} | ||
+ | 16x-6x&=4+16\,,\\[5pt] | ||
+ | 10x&=20\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
- | + | Divide both sides by 10 to get the answer | |
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- | + | {{Displayed math||<math>x=\frac{20}{10}=2\,\textrm{.}</math>}} | |
- | <math>\ | + | |
+ | Finally, we check that <math>x=2</math> satisfies the equation in the exercise | ||
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>6\ | + | \text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt] |
+ | \text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.} | ||
+ | \end{align}</math>}} |
Version vom 14:17, 23. Sep. 2008
We can simplify the left-hand side in the equation by expanding the squares using the squaring rule
Thus, the equation is
Now, move all x's to the left-hand side (subtract 6x from both sides) and the constants to the right-hand side (add 16 to both sides)
Divide both sides by 10 to get the answer
Finally, we check that \displaystyle x=2 satisfies the equation in the exercise